24 questions across ratio, direct and inverse proportion, compound measures and area and volume scale factors.
⚗️ Ratio and proportion mastery — Higher
Ratio, proportion and rates of change is 20% of a Higher paper and it is the area that rewards method over recall. This sheet works through it in one sitting: simplifying and sharing in a ratio, including the questions that give you one share or the difference rather than the total; expressing one quantity as a fraction or percentage of another; direct and inverse proportion, both numerically and as an equation with a constant k; compound measures — speed, density and pressure — with unit conversions built in rather than avoided; and the similar-shapes work where lengths, areas and volumes scale by the factor, its square and its cube. That last group is where Higher candidates lose the most marks in this area, so it is deliberately over-represented here.
- 1.A shade of paint is made by mixing blue paint and white paint. To make 5 litres of the shade, 2 litres of blue paint is used and the rest is white paint. Write the ratio of blue paint to white paint in its simplest form.
- 2.A car travels for 3 hours at an average speed of 80 km/h and then for 1 hour at an average speed of 40 km/h. Work out the average speed for the whole journey.
- 3.Two investors put money into a business in the ratio 3:5. The first investor puts in £1,200. Work out the total amount invested by both investors.
- 4.Write 2 m : 150 cm : 50 cm as a ratio of whole numbers in its simplest form.
- 5.A car is bought for £17,500. Its value decreases by 12% in the first year, and by a further 10% of its reduced value in the second year. Work out the value of the car at the end of the second year, giving your answer to the nearest pound.
- 6.A machine fills bottles at a constant rate. It fills 18 bottles in 3 minutes. Working at the same rate, work out how many bottles the machine fills in 8 minutes.
- 7.Two mathematically similar cylinders have heights in the ratio 3 : 4. Write the ratio of their volumes in its simplest form.
- 8.Two mathematically similar picture frames have perimeters 30 cm and 45 cm. The area of the smaller frame is 40 cm². Work out the area of the larger frame.
- 9.A car uses fuel at a constant rate. It uses 24 litres of fuel to travel 300 km. Assuming the same rate, work out how many litres of fuel are needed for a journey of 175 km.
- 10.The manufacturing cost of a specialist lens, in pounds, is proportional to the square root of its diameter, in millimetres. A lens of diameter 9 mm costs £12 to manufacture. Construct the equation connecting cost C and diameter d, then work out the cost of manufacturing a lens of diameter 16 mm.
- 11.Priya invests £750 in a savings account that pays simple interest. After 3 years, the account contains £840. Work out the annual rate of simple interest.
- 12.A furniture maker builds two mathematically similar bookcases. The smaller bookcase is 40 cm tall and uses 2 m² of wood. The larger bookcase is 60 cm tall. Work out the area of wood needed for the larger bookcase, giving your answer in m².
- 13.Two cars travel at constant speeds. Car A travels 150 km in 3 hours. Car B travels 180 km in 4 hours. Which car is faster, and what is its speed?
- 14.A mortar mix is made from sand and cement in the ratio 5 : 3. Write the ratio of sand to the total mix in its simplest form.
- 15.Jamal invests £600 in a savings account paying 3% simple interest per year. Work out the total amount in the account after 4 years.
- 16.Two quantities y and z are each in direct proportion to x, and are given by y = 3x and z = 7x. Work out the difference between the value of y and the value of z when x = 4.y = 3x
- 17.The strength of a radio signal, in units, is inversely proportional to the square of the distance from the transmitter, in km. At a distance of 2 km the signal strength is 20 units. Construct the equation connecting signal strength S and distance d, then work out the distance at which the signal strength is 5 units.
- 18.y is directly proportional to x², and x is positive. When x = 4, y = 32. Construct the equation connecting x and y, then work out the value of x when y = 200.
- 19.The density of a type of wood is 0.8 g/cm³. Work out the mass of a piece of this wood with a volume of 150 cm³.
- 20.The price of a cycling helmet rises from £80 to £116. Work out the percentage increase.
- 21.y is directly proportional to x. When x = 7, the value of y is 21. Work out the value of x when y = 12.
- 22.The number of members of a running club increases from 45 to 54. Work out the percentage increase.
- 23.Tap A fills a swimming pool in 6 hours. Tap B pours water twice as fast as tap A. The time taken to fill the pool is inversely proportional to the rate of flow. Work out how long tap B takes to fill the pool.
- 24.Order these three values from smallest to largest: 3/8, 0.43, 41%.
Answer key
- (c) 2:3 — The white paint is 5 − 2 = 3 litres. The ratio of blue paint to white paint is 2 : 3, which has no common factor, so it is already in simplest form. Getting 2 : 5 compares the blue paint to the total amount of shade instead of to the white paint. Getting 3 : 2 has the two parts the wrong way round. Getting 5 : 3 uses the total amount of shade instead of the blue paint as the first part.
- (a) 70 km/h — Method: average speed for a whole journey is the total distance divided by the total time, not the mean of the separate speeds. Working: 3 × 80 = 240 km and 1 × 40 = 40 km, giving 280 km in 3 + 1 = 4 hours, so 280 ÷ 4 = 70. Answer: 70 km/h. The distractors: 60 km/h comes from taking the mean of 80 and 40, which would only be right if equal times were spent at each speed; 280 km/h is the total distance written with a speed unit, from forgetting to divide by the total time; 80 km/h comes from quoting the speed of the longer leg as the average for the whole journey.
- (c) £3,200 — Method: find the value of one part of the ratio from the first investor's amount, then work out the second investor's share before adding both together. Working: £1,200 is 3 parts, so one part is £1,200 ÷ 3 = £400. The second investor's share is 5 × £400 = £2,000, and the total is £1,200 + £2,000 = £3,200. So the total invested is £3,200. Distractor £2,000 is only the second investor's share, without adding the first investor's £1,200. Distractor £2,400 comes from doubling the first investor's amount instead of using the ratio. Distractor £6,000 comes from multiplying £1,200 by 5 directly instead of first finding the value of one part.
- (a) 4:3:1 — Convert every part to the same unit: 2 m = 200 cm, so the ratio is 200 : 150 : 50. Dividing all three parts by 50 gives 4 : 3 : 1. Writing 2 : 150 : 50 has not converted 2 m into centimetres, so the units do not match. Writing 3 : 4 : 1 has the first two parts the wrong way round. Writing 4 : 3 : 2 comes from an arithmetic slip on the last part: 50 ÷ 50 = 1, not 2.
- (b) £13,860 — Method: apply the first year's percentage decrease, then apply the second year's percentage decrease to the new value. Working: after the first year, the car is worth £17,500 × 0.88. Multiplying this result by 0.90 gives the value at the end of the second year, £13,860. Answer: £13,860. £13,650 comes from adding the two percentages together (12% + 10% = 22%) and applying a single 22% decrease, £17,500 × 0.78 = £13,650, instead of applying the decreases one after the other. £15,750 comes from applying only the second year's 10% decrease to the original price, forgetting the first year's decrease entirely, £17,500 × 0.90 = £15,750. £15,400 comes from applying only the first year's 12% decrease and stopping there, forgetting to apply the second year's decrease at all.
- (d) 48 — Find the rate first: 18 ÷ 3 = 6 bottles per minute. Then apply it to the new time: 6 × 8 = 48 bottles. Working out 18 + (8 − 3) = 23 adds the extra 5 minutes onto the number of bottles instead of scaling proportionally. Working out 18 × 8 = 144 multiplies the given number of bottles by the new number of minutes without finding the rate first. Writing 18 keeps the count the same, not realising it must change with the time. In 8 minutes the machine fills 48 bottles.
- (c) 27 : 64 — For similar solids, the ratio of volumes is the ratio of lengths cubed: 3³ : 4³ = 27 : 64. 3 : 4 comes from using the height ratio itself as the volume ratio, without cubing it at all. 9 : 16 comes from squaring each part instead of cubing (3² : 4²) — squaring is the rule for area, not volume. 27 : 4 comes from cubing only the first part of the ratio (3³ = 27), and leaving the second part uncubed.
- (c) 90 cm² — The perimeter ratio is 30 : 45, which simplifies to 2 : 3, so the larger frame is 1.5 times the perimeter of the smaller one. Areas scale with the square of this length scale factor: 1.5² = 2.25. 40 × 2.25 = 90, so the larger frame has an area of 90 cm². Giving 60 cm² uses the scale factor, 1.5, without squaring it (40 × 1.5 = 60). Giving 135 cm² cubes the scale factor, 1.5³ = 3.375, as if area scaled like a volume (40 × 3.375 = 135). Giving 2.25 cm² is the squared scale factor on its own, without multiplying by the smaller frame's area of 40 cm².
- (c) 14 litres — Method: find the amount of fuel used per km first, then use it to find the fuel needed for 175 km. Working: 24 ÷ 300 = 0.08 litres per km, and 0.08 × 175 = 14 litres. So 14 litres are needed. Distractor 24 litres comes from assuming the same amount of fuel is used no matter the distance, without scaling. Distractor 21 litres comes from misreading the original distance as 200 km instead of 300 km. Distractor 1.4 litres comes from a decimal-point slip, giving an answer ten times too small.
- (b) £16 — Since cost is proportional to the square root of diameter, C = k√d. Using d = 9, C = 12: √9 = 3, so 12 = k × 3, giving k = 12 ÷ 3 = 4. The equation is C = 4√d. When d = 16: √16 = 4, so C = 4 × 4 = 16. Halving the new diameter instead of taking its square root gives 16 ÷ 2 = 8, and then C = 4 × 8 = 32 — halving a number is not the same as taking its square root, as √16 = 4, not 8. Multiplying k by the diameter itself instead of by its square root gives C = 4 × 16 = 64, skipping the square root altogether. Reporting √16 on its own, without multiplying by k, gives only 4, not the cost. The cost of manufacturing a lens of diameter 16 mm is £16.
- (b) 4% — Method: find the total interest earned, share it equally across the number of years to find one year's interest, then write it as a percentage of the amount invested. Working: total interest = £840 − £750 = £90, so one year's interest is £90 ÷ 3 = £30, and £30 as a percentage of £750 is (£30 ÷ £750) × 100 = 4%. Answer: 4%. 12% comes from treating the total interest of £90 as if it were earned in a single year, (£90 ÷ £750) × 100 = 12%, forgetting to divide by 3 years. 0.04% comes from finding the correct decimal, £30 ÷ £750 = 0.04, but forgetting to multiply by 100 to convert it into a percentage. 112% comes from writing the final amount, £840, as a percentage of the amount invested, £750, without first subtracting the £750 to find the interest alone.
- (b) 4.5 m² — The height ratio is 40 : 60, which simplifies to 2 : 3, so the larger bookcase is 1.5 times as tall as the smaller one. Areas scale with the square of the length scale factor, so the wood needed scales by 1.5² = 2.25. 2 × 2.25 = 4.5, so the larger bookcase needs 4.5 m² of wood. Giving 3 m² uses the length scale factor, 1.5, without squaring it (2 × 1.5 = 3). Giving 6.75 m² cubes the scale factor, 1.5³ = 3.375, as if wood coverage were a volume (2 × 3.375 = 6.75). Giving 2.25 m² is the squared scale factor on its own, without multiplying by the smaller bookcase's wood area of 2 m².
- (b) Car A, 50 km/h — Method: speed = distance ÷ time for each car, then compare. Working: Car A = 150 ÷ 3 = 50 km/h. Car B = 180 ÷ 4 = 45 km/h. Since 50 > 45, Car A is faster, travelling at 50 km/h. Wrong options: Car B, 45 km/h correctly finds Car B's speed but wrongly names the slower car as faster; Car A, 45 km/h picks the correct car but uses Car B's speed by mistake; Car B, 50 km/h picks the wrong car but uses Car A's correct speed value.
- (b) 5:8 — The mix has 5 parts sand and 3 parts cement, so 5 + 3 = 8 parts in total. Sand to total is 5 : 8, and since the highest common factor of 5 and 8 is 1, this is already in its simplest form. Giving 5 : 3 answers sand to cement, not sand to the total mix. Giving 3 : 8 is cement to total, the wrong part of the mix. Giving 8 : 5 has the total and the sand swapped round.
- (c) £672 — Simple interest per year = 3% of £600 = £18. Over 4 years the interest is 18 × 4 = £72. Total in the account = £600 + £72 = £672. A student who gives just the interest, without adding it to the principal, writes £72. A student who adds only one year's interest instead of four gets £600 + £18 = £618. A student who wrongly compounds the interest each year gets 600 × 1.03⁴ = £675.31.
- (b) 16 — Method: substitute x into each equation separately, then subtract the smaller value from the larger. Working: y = 3 × 4 = 12 and z = 7 × 4 = 28, so the difference is 28 − 12 = 16. Answer: 16. The distractors: 4 comes from subtracting the constants, 7 − 3, which is the difference between the two gradients rather than the difference between the values at x = 4; 40 comes from adding the two values, 12 + 28, instead of subtracting them; 28 is the value of z on its own, given instead of being compared with the value of y.
- (c) 4 km — Since signal strength is inversely proportional to the square of the distance, S = k/d². Using d = 2, S = 20: 2² = 4, so 20 = k ÷ 4, giving k = 20 × 4 = 80. The equation is S = 80/d². When S = 5: d² = 80 ÷ 5 = 16, so d = 4 (taking the positive root, since distance cannot be negative). Stopping at d² = 16 without taking the square root leaves 16, the square of the distance, not the distance itself. Treating the relationship as inversely proportional to distance itself, rather than to its square, gives k = 20 × 2 = 40 and then d = 40 ÷ 5 = 8, a different relationship. Multiplying by S instead of dividing by it when isolating d² gives d² = 80 × 5 = 400 and d = 20, the wrong operation. The distance at which the signal strength is 5 units is 4 km.
- (a) 10 — Since y is directly proportional to x², y = kx². Using x = 4, y = 32: 32 = k × 16, so k = 32 ÷ 16 = 2. The equation is y = 2x². When y = 200: x² = 200 ÷ 2 = 100, so x = 10, taking the positive root because the question states that x is positive. Stopping at x² = 100 without taking the square root leaves 100, not the value of x itself. Treating the relationship as if y were proportional to x itself gives k = 32 ÷ 4 = 8 and then x = 200 ÷ 8 = 25, which is a different relationship entirely. Multiplying by k instead of dividing when rearranging gives x² = 200 × 2 = 400 and x = 20, which is not this equation rearranged correctly. The value of x when y = 200 is 10.
- (d) 120 g — Mass = density × volume, so 0.8 × 150 = 120 g. Working out 150 ÷ 0.8 = 187.5 divides by the density instead of multiplying, the wrong way round for finding a mass. Working out 150 × 8 = 1200 misplaces the decimal point in the density, treating 0.8 g/cm³ as 8 g/cm³. Working out 150 − 0.8 = 149.2 simply subtracts the density from the volume, which does not give a mass. The piece of wood has a mass of 120 g.
- (a) 45% — Method: percentage increase = increase ÷ original amount × 100. Working: the increase is £116 − £80 = £36, and 36 ÷ 80 = 0.45, so 0.45 × 100 = 45%. Answer: 45%. The distractors: 36% comes from quoting the £36 increase as though pounds and per cent were the same thing; 31% comes from dividing the £36 increase by the new price £116 instead of by the original £80, which gives 31% to the nearest per cent; 145% is the new price written as a percentage of the original price, which is the whole of the new price rather than the increase.
- (a) 4 — Method: find the constant of proportionality from the pair given, write the equation, then substitute the new value of y and solve. Working: k = 21 ÷ 7 = 3, so y = 3x; putting y = 12 gives 12 = 3x, and x = 12 ÷ 3 = 4. Answer: 4. The distractors: 36 comes from multiplying by the constant instead of dividing by it, 12 × 3, which is the proportion set up upside down; 84 comes from multiplying 12 by the 7 from the first pair, using a value of x as though it were the constant; 9 comes from working out 12 − 3, treating the equation as y = x + 3 rather than y = 3x.
- (c) 20% — Method: percentage increase = (increase ÷ original) × 100. Working: the increase is 54 − 45 = 9, and 9 ÷ 45 = 0.2, so the percentage increase is 0.2 × 100 = 20. Answer: 20%. The distractors: 9% comes from writing the actual increase as a percentage; 16.7% comes from dividing by the new value 54 instead of the original 45; 120% is the multiplier 1.2 written as a change rather than the change itself.
- (a) 3 hours — Method: for a fixed pool the rate of flow multiplied by the time taken is constant, so multiplying the rate by a factor divides the time by that same factor. Working: tap B's rate is 2 times tap A's rate, so tap B's time is 6 ÷ 2 = 3 hours. Answer: 3 hours. The distractors: 12 hours comes from multiplying the time by 2 as well, which treats the time as directly proportional to the rate and has the faster tap taking longer; 4 hours comes from reading ‘twice as fast’ additively, as two hours quicker, and working out 6 − 2 instead of scaling the time by a factor of 2; 1.5 hours comes from applying the factor of 2 twice, halving 6 to 3 and then halving again.
- (a) 3/8, 41%, 0.43 — Method: convert every value to a decimal so they can be compared on the same scale, then order them. Working: 3/8 = 0.375, 41% = 0.41, and 0.43 stays as 0.43, so from smallest to largest the decimals are 0.375, 0.41, 0.43, giving the order 3/8, 41%, 0.43. Answer: 3/8, 41%, 0.43. The order 3/8, 0.43, 41% comes from comparing 0.43 and 41% as raw digits (43 versus 41) without converting 41% into the decimal 0.41 first, wrongly placing 0.43 before 41%. The order 0.43, 41%, 3/8 comes from placing the values in completely reversed order, from largest to smallest instead of smallest to largest. The order 41%, 0.43, 3/8 comes from ordering the values by their TYPE (percentage, then decimal, then fraction) rather than by their actual size.
What is on this worksheet?
The sheet holds 24 questions drawn from the MathsUK bank — the content area covered: Ratio, proportion and rates of change (statements R4, R5, R9, R10, R11, R12, R13). It is pitched at GCSE Higher and takes about 45 minutes to work through in full. It is built for independent practice, with full answers at the end for self-marking.
How to use the sheet well
- Print it or open it on screen — both work. Printing is A4; the screen view fits phones and tablets.
- Do all 24 questions before checking — about 45 minutes is the guide, but there is no time pressure.
- Check the answers — press “Show answers” or print the answer page separately.
- Redo the questions you got wrong — twice as effective as doing 24 fresh ones.
- “New questions” — builds a fresh sheet on the same statements, so you can practise again without repeats.
Why this sheet helps
MathsUK worksheets use questions graded by difficulty and a fair spread of correct-answer positions (the answer is not always (a)) — so the student really has to think about each question rather than guess a pattern. Every question is tagged to a DfE content statement and checked before it enters the bank. The answers come with a step-by-step explanation, not just a value — so a wrong answer becomes a lesson.
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